A function with unexpectedly simple Legendre transformation

Let $$I(x) = \frac{1}{2\pi} \int_{-2}^2 \sqrt{4-y^2}\ln|x-y|dy$$. Then $$I(x)$$ is a concave function and
$$$$I(x)= \begin{cases} \frac{1}{4}x^2-\frac{1}{2}, &\text{if } |x|\leq2 \\ \frac{1}{4}x^2-\frac{1}{2}-\frac{1}{4}|x|\sqrt{x^2-4}+\ln \frac{|x|+\sqrt{x^2-4}}{2} &\text{if } |x|>2 \end{cases}$$$$ Let $$J(x)$$ be the Legendre transform of $$\frac{1}{2}x^2 - I(x)$$, i.e. $$J(x): = \sup_{y\in \mathbb{R}}\{xy - \frac{1}{2}y^2 + I(y)\}$$, then $$J(x)$$ has a quite simple form $$$$J(x) = \begin{cases} x^2-\frac{1}{2} &\text{if } |x|\leq1 \\ \frac{1}{2}x^2 + \ln |x| &\text{if } |x|>1 \end{cases}$$$$ The simple form of $$J(x)$$ suggests some hidden deeper truth. My question is, can we find $$J(x)$$ directly without calculating $$I(x)$$ explicitly?

• for $y>2$, the function $y^2/2-I(y)$ is precisely the rate function for the top eigenvalue of GOE/GUE, up to a constant, see section 6 of link.springer.com/content/pdf/10.1007/PL00008774.pdf. Thus you are asking about the log-mgf of the top eigenvalue, which in turns can be computed using the Selberg integral. I am however not sure this qualifies as a simpler, or deeper, truth.. Commented Jun 13, 2020 at 16:24

Claim. $$J(x)=\tfrac12x^2+\ln|x|+c$$ for $$|x|>1$$ with $$c$$ constant.
Proof: Here, an explicit form for $$I(x)$$ is not needed but I can't use it to prove $$c=0$$.
For $$|x|>1$$, the substitution $$y=2\cos t$$ followed by $$(x-\cos t)(x-\cos s)=x^2-1$$ yields $$I'(2x)=\frac1{2\pi}\int_{-2}^2\frac{\sqrt{4-y^2}}{|2x-y|}\,dy=x-\frac1\pi\int_0^\pi\frac{x^2-1}{|x-\cos t|}dt=x-\sqrt{x^2-1}$$ where $$'=d/dx$$. Thus $$I'(x)+1/I'(x)=x$$ for $$|x|>2$$.
We have $$J(x)=xy_*-\tfrac12y_*^2+I(y_*)$$ where $$x-y_*+I'(y_*)=0$$. This is equivalent to $$y_*-x+\frac1{y_*-x}=y_*$$ using the above identity, which implies $$y_*=x+\frac1x$$. Thus $$J(x)=\frac12x^2-\frac1{2x^2}+I(x+\frac1x)$$ so we now evaluate $$I(x+\frac1x)$$.
Let $$K(x)=xI^*(x+\frac1x)$$ where $$^*=d/d(x+\frac1x)$$ so the above identity simplifies to $$K(x)^2-(x^2+1)K(x)+x^2=0$$. By inspection, we have $$K(x)=1$$ or $$x^2$$; however, if $$K(x)=x^2$$ then $$I'$$ behaves like $$x$$ asymptotically, which is a contradiction. Thus $$K(x)=1$$, so $$I^*(x+\frac1x)=\frac1x$$ implies $$I'(x+\frac1x)=\frac1x(1-\frac1{x^2})$$ or that $$I(x+\frac1x)=\frac1{2x^2}+\ln|x|+c$$. This holds for $$|x|>1$$ so $$J(x)=\tfrac12x^2+\ln|x|+c$$ for $$|x|>1$$.
• $I(2)=1/2$ proves $c=0$ but that requires direct evaluation. Commented Mar 3 at 13:18